In the experiment set up of meter bridge shown in the figure. The null point is obtained at a distance of 40 cm from A . If a $12 Ω$ resistor is connected in series with $R_{1}$ , the null point shifts by 10 cm. The resistance that should be connected in parallel with $(R_{1} + 12) Ω$ such that the null point shifts back to initial position is

Moderate

Solution

$I n i t i a l l y, \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{R_{1}}{R_{2}} = \frac{(40)}{(100 - 40)} = \frac{40}{60} = \frac{2}{3} \to (1) N o w, 12 Ω i s c o n n e c t e d i n s e r i e s w i t h R_{1} . T h i s i n c r e a s e s t h e v a l u e o f r e s i s \tan c e o n l e f t s i d e a n d n u l l p o i n t s h i f t s t o w a r d s r i g h t b y 10 c m . T h e n e w b a l a n c i n g l e n g t h b e c o m e s 50 c m . ∴ \frac{R_{1} + 12}{R_{2}} = \frac{(50)}{(100 - 50)} = 1 \Rightarrow R_{1} + 12 = R_{2} \to (2) U \sin g (1) a n d (2), \frac{2 R_{2}}{3} + 12 = R_{2} \Rightarrow 12 = \frac{R_{2}}{3} \Rightarrow R_{2} = 36 Ω a n d f r o m (1), R_{1} = 24 Ω F i n a l l y, l e t s a y " R " i s c o n n e c t e d i n p a r a l l e l w i t h (R_{1} + 12 = 24 + 12 = 36 Ω) . A s a r e s u l t, n u l l p o i n t s h o u l d s h i f t l e f t w a r d s b y 10 c m m a k i n g b a l a n c i n g l e n g t h a g a i n 40 c m ∴ \frac{R_{e q}}{R_{2}} = \frac{(40)}{(100 - 40)} \Rightarrow \frac{(\frac{36 \times R}{36 + R})}{36} = \frac{2}{3} \Rightarrow R = 72 Ω$

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